Question

An arrow 2.5 cm high is placed at a distance of 25 cm from a diverging mirror of focal length 20 cm. Find the nature, position and size of the image formed.

Answer 1

Given,
The mirror is a diverging mirror, i.e. convex mirror.

Height of the object 'h_o' = 2.5 cm

Distance of the object from the mirror 'u' = -25 cm

Focal length of the convex mirror 'f' = 20 cm

We have to find the position of the image 'v', height of the image 'h_i' and its magnification 'm'.

Using the mirror formula, we get
$$\begin{gathered} \frac{\mathit{1}}{\mathit{f}}\mathit{=}\frac{\mathit{1}}{\mathit{v}}\mathit{+}\frac{\mathit{1}}{\mathit{u}} \\ \frac{\mathit{1}}{\mathit{20}}\mathit{=}\frac{\mathit{1}}{\mathit{v}}\mathit{+}\frac{\mathit{1}}{\mathit{-}\mathit{25}} \\ \frac{1}{20}+\frac{1}{25}=\frac{1}{\mathrm{v}} \\ \frac{5}{100}+\frac{4}{100}=\frac{1}{v} \\ \frac{1}{v}=\frac{9}{100} \\ v=\frac{100}{9}=11.1\mathrm{cm} \end{gathered}$$

Thus, the distance of the image 'v' is 11.1 cm behind the mirror.
Now, using the magnification formula, we get

$$\begin{gathered} m=\frac{-\mathrm{v}}{\mathrm{u}}=\frac{-11.1}{-25} \\ \mathrm{m}=0.44 \end{gathered}$$

Therefore, the image is virtual, erect and smaller in size.
Again,

$$\begin{gathered} m=\frac{h_i}{h_o} \\ 0.4=\frac{h_i}{2.5} \\ {h}_i=0.44\times 2.5=1.1cm \end{gathered}$$
Therefore, the height of the image is 1.1cm.